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There are 5 green 7 red balls. Two balls are selected one by one without replacement. Find the probability that first is green and second is red.

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Let A = first ball is green, B = second ball is red.

We need to find to the probability that first ball is green and second ball is red:

P(A\cap B)=P(A)\cdot P(B|A)

There are 12 balls and 5 of them are green. So, P(A)=\frac{5}{12} .

Since 1 green ball is selected, there are 11 balls and 7 of them are red. So, P(B|A)=\frac{7}{11} .


P(A\cap B)=\frac{5}{12}\cdot \frac{7}{11}=\frac{35}{132} .


Answer: \frac{35}{132} .

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Question ID: mtid-5-stid-8-sqid-1186-qpid-924